Differentiability Questions in English

(C) Given the function $f(x) = \begin{cases} \frac{2 x e^{\frac{1}{2 x}}-3 x e^{\frac{-1}{2 x}}}{e^{\frac{1}{2 x}}+4 e^{-\frac{1}{2 x}}} & x \neq 0 \\ 0 & x=0 \end{cases}$.
First,calculate the Left Hand Derivative ($L$.$H$.$D$.) at $x=0$:
$f^{\prime}(0^{-}) = \lim_{h \rightarrow 0^{+}} \frac{f(0-h)-f(0)}{-h} = \lim_{h \rightarrow 0^{+}} \frac{f(-h)-0}{-h}$
$= \lim_{h \rightarrow 0^{+}} \frac{-2h e^{\frac{-1}{2h}} + 3h e^{\frac{1}{2h}}}{-h(e^{\frac{-1}{2h}} + 4e^{\frac{1}{2h}})} = \lim_{h \rightarrow 0^{+}} \frac{2e^{\frac{-1}{2h}} - 3e^{\frac{1}{2h}}}{e^{\frac{-1}{2h}} + 4e^{\frac{1}{2h}}}$
Divide numerator and denominator by $e^{\frac{1}{2h}}$:
$= \lim_{h \rightarrow 0^{+}} \frac{2e^{\frac{-1}{h}} - 3}{e^{\frac{-1}{h}} + 4} = \frac{0 - 3}{0 + 4} = -\frac{3}{4}$.
Next,calculate the Right Hand Derivative ($R$.$H$.$D$.) at $x=0$:
$f^{\prime}(0^{+}) = \lim_{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0^{+}} \frac{f(h)-0}{h}$
$= \lim_{h \rightarrow 0^{+}} \frac{2h e^{\frac{1}{2h}} - 3h e^{\frac{-1}{2h}}}{h(e^{\frac{1}{2h}} + 4e^{\frac{-1}{2h}})} = \lim_{h \rightarrow 0^{+}} \frac{2e^{\frac{1}{2h}} - 3e^{\frac{-1}{2h}}}{e^{\frac{1}{2h}} + 4e^{\frac{-1}{2h}}}$
Divide numerator and denominator by $e^{\frac{1}{2h}}$:
$= \lim_{h \rightarrow 0^{+}} \frac{2 - 3e^{\frac{-1}{h}}}{1 + 4e^{\frac{-1}{h}}} = \frac{2 - 0}{1 + 0} = 2$.
Since $f^{\prime}(0^{-}) \neq f^{\prime}(0^{+})$,the function $f(x)$ is not differentiable at $x=0$.

(B) Given $f(x) = \begin{cases} 2x+3, & x \leq 1 \\ ax^{2}+bx, & x > 1 \end{cases}$.
Since $f(x)$ is differentiable for all $x \in R$,it must be continuous at $x = 1$.
$\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1) \Rightarrow 2(1) + 3 = a(1)^2 + b(1) \Rightarrow a + b = 5 \quad \dots(i)$
Also,$f'(x)$ must be continuous at $x = 1$.
$f'(x) = \begin{cases} 2, & x < 1 \\ 2ax + b, & x > 1 \end{cases}$
$\lim_{x \to 1^{-}} f'(x) = \lim_{x \to 1^{+}} f'(x) \Rightarrow 2 = 2a(1) + b \Rightarrow 2a + b = 2 \quad \dots(ii)$
Subtracting $(i)$ from $(ii)$: $(2a + b) - (a + b) = 2 - 5 \Rightarrow a = -3$.
Substituting $a = -3$ in $(i)$: $-3 + b = 5 \Rightarrow b = 8$.
Thus,for $x > 1$,$f(x) = -3x^2 + 8x$.
We need to find $f(2)$.
$f(2) = -3(2)^2 + 8(2) = -3(4) + 16 = -12 + 16 = 4$.

(D) The function is defined as $f(x) = |x - 1| + |x - 2|$ for $x \in [0, 3]$.
By breaking the absolute values,we get:
$f(x) = \begin{cases} -(x-1) - (x-2) = -2x + 3, & 0 \leq x \leq 1 \\ (x-1) - (x-2) = 1, & 1 < x < 2 \\ (x-1) + (x-2) = 2x - 3, & 2 \leq x \leq 3 \end{cases}$
Since $f(x)$ is a sum of continuous functions,it is continuous everywhere in $[0, 3]$.
At $x = 1$,the left-hand derivative is $\frac{d}{dx}(-2x+3) = -2$ and the right-hand derivative is $\frac{d}{dx}(1) = 0$. Since $-2 \neq 0$,$f(x)$ is not differentiable at $x = 1$.
At $x = 2$,the left-hand derivative is $\frac{d}{dx}(1) = 0$ and the right-hand derivative is $\frac{d}{dx}(2x-3) = 2$. Since $0 \neq 2$,$f(x)$ is not differentiable at $x = 2$.
Thus,$f(x)$ is continuous but not differentiable at $x = 1$ and $x = 2$.

(D) Given $f(x) = |(|x| + \alpha)(|x| - 1)|$.
Let $g(x) = (|x| + \alpha)(|x| - 1)$.
The function $f(x) = |g(x)|$ is not differentiable at the points where $g(x) = 0$ (provided the roots are simple) and at $x = 0$ due to the $|x|$ term.
Case $1$: If $\alpha > 0$,then $|x| + \alpha = 0$ has no real solution. The roots of $g(x) = 0$ are $|x| = 1$,i.e.,$x = 1, -1$.
At $x = 1, -1$,the function $f(x)$ has sharp corners. Also,at $x = 0$,$f(x) = |\alpha(-1)| = |-\alpha| = \alpha$. Since $f'(0^+)$ and $f'(0^-)$ will differ due to the $|x|$ term,$f(x)$ is not differentiable at $x = 0$. Thus,for $\alpha > 0$,there are $3$ points of non-differentiability $(x = -1, 0, 1)$.
Case $2$: If $\alpha < 0$ and $\alpha \neq -1$,let $\alpha = -k$ where $k > 0$ and $k \neq 1$. Then $g(x) = (|x| - k)(|x| - 1)$.
The roots are $|x| = k$ and $|x| = 1$,which gives $x = \pm k$ and $x = \pm 1$.
These are $4$ distinct points where $f(x) = 0$. Additionally,$f(x)$ is not differentiable at $x = 0$ because of the $|x|$ term.
Thus,for $\alpha < 0$ (and $\alpha \neq -1$),there are $5$ points of non-differentiability $(x = -1, -k, 0, k, 1)$.
Comparing with the options,$D$ is correct.

(D) To check the differentiability of $f(x)$ at $x = a$,we calculate the left-hand derivative ($L$.$H$.$D$.) and right-hand derivative ($R$.$H$.$D$.).
$L$.$H$.$D$. at $x = a$ is $\lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} = \lim_{h \to 0} \frac{\frac{1}{2}(b^2 - a^2) - \frac{1}{2}(b^2 - a^2)}{-h} = 0$.
Now,$R$.$H$.$D$. at $x = a$ is $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{\frac{1}{2}b^2 - \frac{(a+h)^2}{6} - \frac{a^3}{3(a+h)} - \frac{1}{2}(b^2 - a^2)}{h}$.
Simplifying the expression: $\lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{2}a^2 - \frac{(a+h)^3 + 2a^3}{6(a+h)} \right]$.
As $h \to 0$,the expression does not converge to a finite value,implying the derivative does not exist at $x = a$.
Since $f(x)$ is not differentiable at $x = a$,$f'(x)$ is also not differentiable at $x = a$.

(A) fundamental theorem in calculus states that if a function $f(x)$ is differentiable at a point $x=a$,then it must be continuous at that point $x=a$.
The contrapositive of this statement is: if $f(x)$ is not continuous at $x=a$,then $f(x)$ is not differentiable at $x=a$.
Since the Reason $(R)$ provides the direct theorem that justifies the Assertion $(A)$,both are true and $(R)$ is the correct explanation of $(A)$.

(C) The function is defined as $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$.
To find the points of non-differentiability,we analyze the intersections of the functions $y_1 = 3 - x$,$y_2 = 3 + x$,and $y_3 = 6$.
$1$. Intersection of $y_1$ and $y_3$: $3 - x = 6 \implies x = -3$.
$2$. Intersection of $y_2$ and $y_3$: $3 + x = 6 \implies x = 3$.
$3$. Intersection of $y_1$ and $y_2$: $3 - x = 3 + x \implies 2x = 0 \implies x = 0$. At $x=0$,$y_1 = 3$ and $y_2 = 3$,but $y_3 = 6$,so $f(0) = 6$.
The function $f(x)$ is given by:
$f(x) = \begin{cases} 3 + x, & x < -3 \\ 6, & -3 \le x \le 3 \\ 3 - x, & x > 3 \end{cases}$ (Wait,checking the max: for $x < -3$,$3-x > 6$,so $f(x) = 3-x$. For $x > 3$,$3+x > 6$,so $f(x) = 3+x$.)
Correct definition: $f(x) = \begin{cases} 3 - x, & x < -3 \\ 6, & -3 \le x \le 3 \\ 3 + x, & x > 3 \end{cases}$.
The function has sharp corners (points of non-differentiability) at $x = -3$ and $x = 3$.
Thus,$a = -3$ and $b = 3$.
Therefore,$|a| + |b| = |-3| + |3| = 3 + 3 = 6$.

(B) Given,$f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \\ ax^2 + b, & -1 < x < 1 \end{cases}$
Expanding the definition,we get:
$f(x) = \begin{cases} -\frac{1}{x}, & x \leq -1 \\ ax^2 + b, & -1 < x < 1 \\ \frac{1}{x}, & x \geq 1 \end{cases}$
For $f(x)$ to be differentiable $\forall x \in \mathbb{R}$,it must be continuous and differentiable at $x = 1$.
Continuity at $x = 1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\implies a(1)^2 + b = \frac{1}{1} \implies a + b = 1$ . . . $(1)$
Differentiability at $x = 1$: $f'(1^-) = f'(1^+)$.
For $x < 1$,$f'(x) = 2ax$. For $x > 1$,$f'(x) = -\frac{1}{x^2}$.
$\implies 2a(1) = -\frac{1}{(1)^2} \implies 2a = -1 \implies a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into equation $(1)$:
$-\frac{1}{2} + b = 1 \implies b = 1 + \frac{1}{2} = \frac{3}{2}$.
Thus,$a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

(C) To check for continuity and differentiability at $x = 0$,we first evaluate the limit $\lim_{x \to 0} f(x)$.
$\lim_{x \to 0} \frac{x^2 \ln \cos x}{\ln(1 + x^2)} = \lim_{x \to 0} \left( \frac{x^2}{\ln(1 + x^2)} \right) \times \ln \cos x$.
Since $\lim_{x \to 0} \frac{x^2}{\ln(1 + x^2)} = 1$ and $\lim_{x \to 0} \ln \cos x = \ln(1) = 0$,the limit is $1 \times 0 = 0$.
Since $\lim_{x \to 0} f(x) = f(0) = 0$,the function is continuous at $x = 0$.
Now,check for differentiability at $x = 0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \ln \cos h}{h \ln(1 + h^2)} = \lim_{h \to 0} \frac{h \ln \cos h}{\ln(1 + h^2)}$.
Using the standard limits $\lim_{h \to 0} \frac{\ln(1 + h^2)}{h^2} = 1$ and $\lim_{h \to 0} \frac{\ln \cos h}{h^2} = -\frac{1}{2}$:
$f'(0) = \lim_{h \to 0} \left( \frac{h^2}{\ln(1 + h^2)} \right) \times \left( \frac{\ln \cos h}{h^2} \right) \times h = 1 \times (-\frac{1}{2}) \times 0 = 0$.
Since the limit exists and is finite,$f(x)$ is differentiable at $x = 0$.

(D) function $f(x)$ is differentiable at $x=0$ if $\lim_{h \to 0^+} \frac{f(h)-f(0)}{h} = \lim_{h \to 0^-} \frac{f(h)-f(0)}{h}$.
For $f(x) = \sin |x| - |x|$,we have $f(0) = \sin(0) - 0 = 0$.
Right Hand Derivative $(RHD)$ at $x=0$:
$\lim_{h \to 0^+} \frac{\sin |h| - |h| - 0}{h} = \lim_{h \to 0^+} \frac{\sin h - h}{h} = \lim_{h \to 0^+} (\frac{\sin h}{h} - 1) = 1 - 1 = 0$.
Left Hand Derivative $(LHD)$ at $x=0$:
$\lim_{h \to 0^-} \frac{\sin |-h| - |-h| - 0}{h} = \lim_{h \to 0^-} \frac{\sin(-h) - (-h)}{h} = \lim_{h \to 0^-} \frac{-\sin h + h}{h} = \lim_{h \to 0^-} (-\frac{\sin h}{h} + 1) = -1 + 1 = 0$.
Since $LHD = RHD = 0$,the function $f(x) = \sin |x| - |x|$ is differentiable at $x=0$.

(C) To check for differentiability,we examine the points of transition: $x = -\sqrt{5}$ and $x = \sqrt{5}$.
At $x = -\sqrt{5}$:
Left-hand limit $(LHL)$ = $4$.
Right-hand limit $(RHL)$ = $(-\sqrt{5})^2 - 1 = 5 - 1 = 4$.
Since $LHL = RHL = f(-\sqrt{5})$,the function is continuous at $x = -\sqrt{5}$.
Left-hand derivative $(LHD)$ = $\frac{d}{dx}(4) = 0$.
Right-hand derivative $(RHD)$ = $\frac{d}{dx}(x^2-1) = 2x = 2(-\sqrt{5}) = -2\sqrt{5}$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = -\sqrt{5}$.
At $x = \sqrt{5}$:
$LHL$ = $(\sqrt{5})^2 - 1 = 4$.
$RHL$ = $4$.
Since $LHL = RHL = f(\sqrt{5})$,the function is continuous at $x = \sqrt{5}$.
$LHD$ = $2x = 2(\sqrt{5}) = 2\sqrt{5}$.
$RHD$ = $\frac{d}{dx}(4) = 0$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x = \sqrt{5}$.
Thus,there are $k = 2$ points where the function is not differentiable.
Therefore,$k - 2 = 2 - 2 = 0$.

(A) The function $f(x) = |x|$ is defined as:
$f(x) = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$
At $x = 0$,the left-hand derivative is $\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1$.
The right-hand derivative is $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1$.
Since the left-hand derivative $\neq$ right-hand derivative,$f(x)$ is not differentiable at $x = 0$.
However,$f(x)$ is continuous everywhere,including $x = 0$.
For any $x = a \neq 0$,the function is locally linear (either $x$ or $-x$),so it is differentiable.
Thus,Assertion $(A)$ is correct.
Reason $(R)$ states a fundamental theorem in calculus: differentiability implies continuity,but continuity does not imply differentiability. This theorem explains why $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there.
Therefore,$R$ is the correct explanation of $A$.

(A) The function is given by $f(x) = 2x|x|$.
We can rewrite this as a piecewise function:
$f(x) = \begin{cases} 2x^2, & x \geq 0 \\ -2x^2, & x < 0 \end{cases}$
Since $2x^2$ and $-2x^2$ are polynomials,$f(x)$ is differentiable for all $x \neq 0$. We only need to check the differentiability at $x = 0$.
Left Hand Derivative $(LHD)$ at $x = 0$:
$f'(0^-) = \lim_{h \rightarrow 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0^-} \frac{-2h^2 - 0}{h} = \lim_{h \rightarrow 0^-} (-2h) = 0$.
Right Hand Derivative $(RHD)$ at $x = 0$:
$f'(0^+) = \lim_{h \rightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0^+} \frac{2h^2 - 0}{h} = \lim_{h \rightarrow 0^+} (2h) = 0$.
Since $LHD = RHD = 0$,the function $f(x)$ is differentiable at $x = 0$.
Therefore,$f(x)$ is differentiable for all $x \in (-\infty, \infty)$.

(D) The function $y = \sin^{-1}(\cos x)$ is defined when $-1 \le \cos x \le 1$. However,the derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$,which is undefined when $u = \pm 1$.
Thus,the function $y = \sin^{-1}(\cos x)$ is not differentiable where $\cos x = 1$ or $\cos x = -1$.
For $\cos x = -1$,$x = (2n+1)\pi$ for any integer $n$. For $n=0$,$x = \pi$.
For $\cos x = 1$,$x = 2n\pi$ for any integer $n$. For $n=1$,$x = 2\pi$,and for $n=-1$,$x = -2\pi$.
Since the function is not differentiable at $x = \pi$,$x = 2\pi$,and $x = -2\pi$,all the given options are correct.

(D) The function is defined as $f(x) = |\log_e |x||$.
First,consider the domain of the function. The function is undefined at $x = 0$ because $\log_e 0$ is undefined.
Next,examine the points where the function might not be differentiable. The function $f(x)$ involves an absolute value of a logarithm,which creates sharp corners (cusps) where the argument of the absolute value is zero.
Setting $\log_e |x| = 0$,we get $|x| = 1$,which implies $x = 1$ or $x = -1$.
At $x = 1$ and $x = -1$,the graph has sharp turns,meaning the derivative does not exist at these points.
At $x = 0$,the function has a vertical asymptote,so it is not continuous,and therefore not differentiable.
Thus,the function $f(x)$ is differentiable for all real numbers except $x = 0, 1, -1$.
This is represented as $R - \{0, 1, -1\}$ or $R - \{0, \pm 1\}$.
Therefore,option $(d)$ is correct.

(D) The function is defined as $f(x) = [x] + |1 - x|$ on the interval $[-1, 3]$.
Points of discontinuity for the greatest integer function $[x]$ in the interval $[-1, 3]$ are $x = 0, 1, 2, 3$.
At $x = 1$,the function is $f(x) = [x] + |1 - x|$. Since $|1 - x|$ is continuous everywhere,the non-differentiability arises from the jump discontinuities of $[x]$ at $x = 0, 1, 2, 3$ and the corner point of $|1 - x|$ at $x = 1$.
Checking the points:
$1$. At $x = 0$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
$2$. At $x = 1$: $[x]$ is discontinuous and $|1 - x|$ has a corner,so $f(x)$ is not differentiable.
$3$. At $x = 2$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
$4$. At $x = 3$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
Thus,the function is not differentiable at $x = 0, 1, 2, 3$. There are $4$ such points.

(C) Given the function $f(x) = |x - 0.5| + |x - 1| + \tan x$.
We know that the modulus function $|x - a|$ is not differentiable at $x = a$.
Therefore,$|x - 0.5|$ is not differentiable at $x = 0.5$ and $|x - 1|$ is not differentiable at $x = 1$.
Both $x = 0.5$ and $x = 1$ lie within the interval $(0, 2)$.
Additionally,the function $\tan x$ is not defined (and hence not differentiable) at $x = \frac{\pi}{2}$.
Since $\pi \approx 3.14$,we have $\frac{\pi}{2} \approx 1.57$,which also lies within the interval $(0, 2)$.
Thus,the function $f(x)$ is not differentiable at $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$.
There are $3$ such points in the interval $(0, 2)$.
Therefore,option $C$ is correct.

(C) For $f(x)$ to be differentiable everywhere,it must be continuous and differentiable at $x = 1$ and $x = -1$. Since $f(x)$ is an even function,we check $x = 1$.
Continuity at $x = 1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} (\alpha x^2 - \beta) = \alpha - \beta$.
$\lim_{x \to 1^+} (-\frac{1}{x}) = -1$.
So,$\alpha - \beta = -1$ (Equation $1$).
Differentiability at $x = 1$: $f'(1^-) = f'(1^+)$.
$f'(x) = 2\alpha x$ for $|x| < 1$ and $f'(x) = \frac{1}{x^2}$ for $|x| > 1$.
$f'(1^-) = 2\alpha(1) = 2\alpha$.
$f'(1^+) = \frac{1}{(1)^2} = 1$.
Thus,$2\alpha = 1 \implies \alpha = \frac{1}{2}$.
Substituting $\alpha = \frac{1}{2}$ into Equation $1$: $\frac{1}{2} - \beta = -1 \implies \beta = \frac{3}{2}$.
Therefore,the ordered pair $(\alpha, \beta) = (\frac{1}{2}, \frac{3}{2})$.

(B) Given: $f(x) = \frac{|x|}{1 - |x|}$.
First,check continuity at $x = 0$:
$\text{LHL} = \lim_{x \to 0^-} \frac{-x}{1 - (-x)} = \lim_{x \to 0^-} \frac{-x}{1 + x} = 0$.
$\text{RHL} = \lim_{x \to 0^+} \frac{x}{1 - x} = 0$.
Since $f(0) = 0$,$f(x)$ is continuous at $x = 0$.
Now,check differentiability at $x = 0$:
For $x < 0$,$f(x) = \frac{-x}{1 + x}$,so $f'(x) = \frac{-(1+x) - (-x)(1)}{(1+x)^2} = \frac{-1}{(1+x)^2}$. Thus,$\text{LHD} = f'(0^-) = -1$.
For $x > 0$,$f(x) = \frac{x}{1 - x}$,so $f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$. Thus,$\text{RHD} = f'(0^+) = 1$.
Since $\text{LHD} \neq \text{RHD}$,$f(x)$ is not differentiable at $x = 0$.
However,$f(x)$ is differentiable for all $x \in D \setminus \{0\}$.
Therefore,$f$ is differentiable on $D$ except at $x = 0$.

(D) function $f(x)$ is not differentiable at points where the expression inside the absolute value is zero,or where the function itself is discontinuous.
$1$. The absolute value functions $|x - 0.5|$ and $|x - 1|$ are not differentiable at $x = 0.5$ and $x = 1$ respectively.
$2$. The function $\tan x$ is not defined (and thus not differentiable) at $x = \frac{\pi}{2} + n\pi$. Within the interval $(0, 2)$,$\frac{\pi}{2} \approx 1.57$,which lies in the interval.
$3$. Combining these,the function $f(x)$ is not differentiable at $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$.
Thus,the correct option is $D$.

(A) Given the function,$f(x) = \begin{cases} x^2 + bx + c, & x < 1 \\ x, & x \geq 1 \end{cases}$.
For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$ and the left-hand derivative must equal the right-hand derivative.
First,for continuity at $x = 1$:
$\lim_{x \to 1^-} f(x) = f(1)$
$\lim_{x \to 1^-} (x^2 + bx + c) = 1$
$1 + b + c = 1 \Rightarrow b + c = 0$ (Equation $1$).
Next,for differentiability at $x = 1$,the derivatives must match:
$f'(x) = \begin{cases} 2x + b, & x < 1 \\ 1, & x > 1 \end{cases}$
$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x)$
$2(1) + b = 1 \Rightarrow 2 + b = 1 \Rightarrow b = -1$.
Substituting $b = -1$ into Equation $1$:
$-1 + c = 0 \Rightarrow c = 1$.
Therefore,$b - c = -1 - 1 = -2$.

(C) Given $f(x) = |x| + |sin x|$.
For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,the left hand derivative $(LHD)$ at $x=0$ is defined as:
$LHD = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h}$
Since $f(0) = |0| + |sin 0| = 0$,we have:
$LHD = \lim_{h \to 0^+} \frac{|-h| + |sin(-h)| - 0}{-h}$
For small $h > 0$,$|-h| = h$ and $|sin(-h)| = |-sin h| = sin h$ (since $sin h > 0$ for $h \in (0, \pi/2)$).
$LHD = \lim_{h \to 0^+} \frac{h + sin h}{-h}$
$LHD = \lim_{h \to 0^+} -\left( \frac{h}{h} + \frac{sin h}{h} \right)$
$LHD = -(1 + 1) = -2$.

(C) Given the function $f(x) = \frac{x}{1+|x|}$.
We can write this as a piecewise function:
If $x \ge 0$,then $|x| = x$,so $f(x) = \frac{x}{1+x}$.
If $x < 0$,then $|x| = -x$,so $f(x) = \frac{x}{1-x}$.
Now,we find the derivative $f'(x)$:
For $x > 0$,$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$.
For $x < 0$,$f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}$.
At $x = 0$,we check the left-hand derivative and right-hand derivative:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h \to 0^-} \frac{1}{1-h} = 1$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{1+h} = 1$.
Since $LHD = RHD = 1$,the function is differentiable at $x = 0$ and $f'(0) = 1$.
Thus,the derivative exists for all real numbers $x \in (-\infty, \infty)$.

(B) Given $f(x) = \frac{x-1}{2x^2-7x+5}$ for $x \neq 1$.
Factoring the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.
Thus,for $x \neq 1$,$f(x) = \frac{x-1}{(2x-5)(x-1)} = \frac{1}{2x-5}$.
The function is $f(x) = \begin{cases} \frac{1}{2x-5}, & x \neq 1 \\ -\frac{1}{3}, & x=1 \end{cases}$.
By the definition of the derivative,$f^{\prime}(1) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{\frac{1}{2(1+h)-5} - (-\frac{1}{3})}{h} = \lim_{h \to 0} \frac{\frac{1}{2h-3} + \frac{1}{3}}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{3 + (2h-3)}{3h(2h-3)} = \lim_{h \to 0} \frac{2h}{3h(2h-3)}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{2}{3(2h-3)} = \frac{2}{3(-3)} = -\frac{2}{9}$.

(C) Given $f(x) = \max\{\cos x, \sin x, 0\}$.
In the interval $[0, 2\pi]$,the function $f(x)$ is defined as:
$f(x) = \cos x$ for $x \in [0, \pi/4]$
$f(x) = \sin x$ for $x \in [\pi/4, 5\pi/4]$
$f(x) = 0$ for $x \in [5\pi/4, 2\pi]$
Checking for non-differentiability at the intersection points:
$1$. At $x = \pi/4$,$\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$. The derivatives are $-\sin(\pi/4) = -1/\sqrt{2}$ and $\cos(\pi/4) = 1/\sqrt{2}$. Since they are not equal,$f(x)$ is not differentiable at $x = \pi/4$.
$2$. At $x = 5\pi/4$,$\sin(5\pi/4) = -1/\sqrt{2}$ and $0$. Since $\sin(5\pi/4) < 0$,the function is $0$. The transition from $\sin x$ to $0$ occurs at $x = \pi$ (where $\sin x = 0$). At $x = \pi$,$\sin(\pi) = 0$ and the derivative is $\cos(\pi) = -1$. The derivative of $0$ is $0$. Since $-1 \neq 0$,it is not differentiable at $x = \pi$.
$3$. At $x = 2\pi$,$\cos(2\pi) = 1$ and $0$. The transition from $0$ to $\cos x$ occurs at $x = 2\pi - \pi/2 = 3\pi/2$ is not correct,actually $\cos x = 0$ at $x = 3\pi/2$. The function $f(x)$ is $0$ on $[5\pi/4, 2\pi]$ and $\cos x$ starts being positive at $x = 3\pi/2$. At $x = 3\pi/2$,$\cos(3\pi/2) = 0$ and the derivative is $-\sin(3\pi/2) = 1$. Since $1 \neq 0$,it is not differentiable at $x = 3\pi/2$.
Thus,there are $3$ points of non-differentiability in every interval of length $2\pi$.
In $(0, 2024\pi)$,there are $1012$ such intervals.
Total points of non-differentiability $= 3 \times 1012 = 3036$.
Given $1012k = 3036$,we get $k = 3$.

(B) For the function $g(x)$ to be differentiable at $x=3$,it must first be continuous at $x=3$.
Continuity at $x=3$: $\lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) = g(3)$.
$K \sqrt{3+1} = 3m+2 \implies 2K = 3m+2$ (Equation $1$).
Differentiability at $x=3$: The left-hand derivative must equal the right-hand derivative.
$g'(x) = \begin{cases} \frac{K}{2\sqrt{x+1}} &, 0 < x < 3 \\ m &, 3 < x < 5 \end{cases}$.
At $x=3$,$\frac{K}{2\sqrt{3+1}} = m \implies \frac{K}{4} = m \implies K = 4m$ (Equation $2$).
Substitute $K=4m$ into Equation $1$: $2(4m) = 3m+2 \implies 8m = 3m+2 \implies 5m = 2 \implies m = \frac{2}{5}$.
Then $K = 4(\frac{2}{5}) = \frac{8}{5}$.
Therefore,$K+m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$.

(C) We check the continuity at $x = \frac{5}{3}$:
Left-hand limit: $\lim_{x \to \frac{5}{3}^-} f(x) = 5 - 3(\frac{5}{3}) = 5 - 5 = 0$.
Right-hand limit: $\lim_{x \to \frac{5}{3}^+} f(x) = (\frac{5}{3})^2 - 3(\frac{5}{3}) + 20 = \frac{25}{9} - 5 + 20 = \frac{25}{9} + 15 = \frac{25 + 135}{9} = \frac{160}{9}$.
Since $\lim_{x \to \frac{5}{3}^-} f(x) \neq \lim_{x \to \frac{5}{3}^+} f(x)$,the function is discontinuous at $x = \frac{5}{3}$.
Now check differentiability at $x = 2$. Since $2 > \frac{5}{3}$,the function is defined by $f(x) = x^2 - 3x + 20$ in the neighborhood of $x = 2$.
This is a polynomial function,which is differentiable everywhere in its domain. Thus,$f(x)$ is differentiable at $x = 2$.

(C) First,we check for continuity at $x=0$: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^2 \ln \cos x}{\ln (1+x^2)}$.
Using standard limits $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$,we have $\lim_{x \to 0} \frac{\ln(1+x^2)}{x^2} = 1$.
Thus,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{x^2}{\ln(1+x^2)} \right) \cdot \ln \cos x = 1 \cdot \ln(1) = 0$.
Since $\lim_{x \to 0} f(x) = f(0) = 0$,the function is continuous at $x=0$.
Now,we check for differentiability at $x=0$ using the definition $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \ln \cos h}{h \ln(1+h^2)} = \lim_{h \to 0} \frac{h \ln \cos h}{\ln(1+h^2)}$.
Dividing numerator and denominator by $h^2$: $\lim_{h \to 0} \frac{\frac{\ln \cos h}{h}}{\frac{\ln(1+h^2)}{h^2}} = \lim_{h \to 0} \frac{\frac{\ln \cos h}{h}}{1}$.
Using $L$'Hospital's rule on $\frac{\ln \cos h}{h}$: $\lim_{h \to 0} \frac{-\tan h}{1} = 0$.
Since the limit exists and is equal to $0$,$f(x)$ is differentiable at $x=0$.

(C) function $f(x) = |g(x)|$ is not differentiable at points where $g(x) = 0$,provided $g(x)$ is a linear or polynomial function.
Given $f(x) = 7|2x + 1| - 19|3x - 5|$.
The function involves absolute value terms $|2x + 1|$ and $|3x - 5|$.
The term $|2x + 1|$ is not differentiable at $2x + 1 = 0$,which gives $x = -\frac{1}{2}$.
The term $|3x - 5|$ is not differentiable at $3x - 5 = 0$,which gives $x = \frac{5}{3}$.
Since the function is a linear combination of these absolute value functions,it is not differentiable at the points where the expressions inside the absolute values are zero.
Therefore,the function $f(x)$ is not differentiable at $x = -\frac{1}{2}$ and $x = \frac{5}{3}$.

(A) The function is given by $f(x) = ||x| - 1|$.
We know that $|x|$ is not differentiable at $x = 0$.
Also,the function $g(x) = |x| - 1$ is not differentiable at $x = 0$.
The function $f(x) = |g(x)|$ is not differentiable where $g(x) = 0$ or where $g(x)$ is not differentiable.
Setting $g(x) = 0$,we get $|x| - 1 = 0$,which implies $|x| = 1$,so $x = 1$ or $x = -1$.
Additionally,$g(x)$ is not differentiable at $x = 0$.
Therefore,$f(x)$ is not differentiable at $x \in \{-1, 0, 1\}$.
Thus,the set of all values of $x$ for which $f(x)$ is differentiable is $R - \{-1, 0, 1\}$.

(A) Since $f(x)$ is differentiable at $x=0$,it must be continuous at $x=0$.
$\lim_{x \rightarrow 0} f(x) = f(0) = P$.
$\lim_{x \rightarrow 0} \frac{(e^{kx}-1) \sin kx}{4 \tan x} = \lim_{x \rightarrow 0} \frac{(e^{kx}-1)}{x} \cdot \frac{\sin kx}{x} \cdot \frac{x^2}{4 \tan x} = (k) \cdot (k) \cdot (0) = 0$.
Thus,$P = 0$.
Now,$f^{\prime}(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x} = \lim_{x \rightarrow 0} \frac{(e^{kx}-1) \sin kx}{4x \tan x}$.
$f^{\prime}(0) = \lim_{x \rightarrow 0} \left( \frac{e^{kx}-1}{x} \right) \left( \frac{\sin kx}{x} \right) \left( \frac{x}{\tan x} \right) \cdot \frac{1}{4} = (k) \cdot (k) \cdot (1) \cdot \frac{1}{4} = \frac{k^2}{4}$.

(B) Statement $(a)$ is incorrect because if a function is differentiable at a point,it must be continuous at that point.
Statement $(b)$ is correct because differentiability implies continuity; therefore,the contrapositive (not continuous implies not differentiable) is also true.
Statement $(c)$ is correct because $f(x) = |x|$ is continuous for all $x \in R$ but is not differentiable at $x = 0$.
Statement $(d)$ is incorrect because $f(x) = x - [x]$ is the fractional part function,which is discontinuous at all integers,including $x = 1$. Since it is discontinuous at $x = 1$,it is not differentiable at $x = 1$.
Thus,statements $(b)$ and $(c)$ are correct.

(B) First,we determine $|g(x)|$:
$|g(x)| = \begin{cases} |-1| = 1, & -2 \le x < 0 \\ |x^2 - 1|, & 0 \le x \le 2 \end{cases} = \begin{cases} 1, & -2 \le x < 0 \\ 1 - x^2, & 0 \le x < 1 \\ x^2 - 1, & 1 \le x \le 2 \end{cases}$
Next,we determine $g(|x|)$:
Since $|x| \ge 0$ for all $x \in [-2, 2]$,we use the second part of $g(x)$:
$g(|x|) = |x|^2 - 1 = x^2 - 1$ for all $x \in [-2, 2]$.
Now,calculate $f(x) = |g(x)| + g(|x|) + 2$:
For $-2 \le x < 0$: $f(x) = 1 + (x^2 - 1) + 2 = x^2 + 2$.
For $0 \le x < 1$: $f(x) = (1 - x^2) + (x^2 - 1) + 2 = 2$.
For $1 \le x \le 2$: $f(x) = (x^2 - 1) + (x^2 - 1) + 2 = 2x^2$.
Thus,$f(x) = \begin{cases} x^2 + 2, & -2 \le x < 0 \\ 2, & 0 \le x < 1 \\ 2x^2, & 1 \le x \le 2 \end{cases}$.
Checking differentiability at $x = 0$:
$LHD = \lim_{x \to 0^-} \frac{d}{dx}(x^2 + 2) = 2(0) = 0$.
$RHD = \lim_{x \to 0^+} \frac{d}{dx}(2) = 0$.
Since $LHD = RHD$,$f$ is differentiable at $x = 0$.
Checking differentiability at $x = 1$:
$LHD = \lim_{x \to 1^-} \frac{d}{dx}(2) = 0$.
$RHD = \lim_{x \to 1^+} \frac{d}{dx}(2x^2) = 4(1) = 4$.
Since $LHD \neq RHD$,$f$ is not differentiable at $x = 1$.

(B) Given the function $f(x) = |x+1| + |x-1|$.
We know that the function $g(x) = |x|$ is not differentiable at $x = 0$.
Similarly,$|x+1|$ is not differentiable at $x = -1$ and $|x-1|$ is not differentiable at $x = 1$.
The sum of two functions is not differentiable at the points where either of the functions is not differentiable.
Therefore,$f(x)$ is not differentiable at $x = -1$ and $x = 1$.

(A) Given $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$.
For $x > 1$,we have $f(x) = 2(x - [x]) - 1$. Since for $x$ slightly greater than $1$,$[x] = 1$,we get $f(x) = 2(x - 1) - 1 = 2x - 3$.
The right-hand derivative at $x = 1$ is defined as $f'(1^+) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$.
First,$f(1) = [\cos \pi] = [-1] = -1$.
Then,$f'(1^+) = \lim_{h \to 0^+} \frac{2(1+h) - 3 - (-1)}{h} = \lim_{h \to 0^+} \frac{2 + 2h - 3 + 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Thus,the right derivative at $x = 1$ is $2$.

(C) We have,$f(x)=\frac{2x}{4+3|x|}, x \in R$.
This can be written as:
$f(x) = \begin{cases} \frac{2x}{4+3x}, & \text{if } x \geq 0 \\ \frac{2x}{4-3x}, & \text{if } x < 0 \end{cases}$
To find $f^{\prime}(0)$,we check the left-hand and right-hand derivatives at $x=0$.
For $x > 0$,$f^{\prime}(x) = \frac{d}{dx} \left( \frac{2x}{4+3x} \right) = \frac{(4+3x)(2) - 2x(3)}{(4+3x)^2} = \frac{8+6x-6x}{(4+3x)^2} = \frac{8}{(4+3x)^2}$.
Thus,$Rf^{\prime}(0) = \lim_{x \rightarrow 0^+} \frac{8}{(4+3x)^2} = \frac{8}{16} = \frac{1}{2}$.
For $x < 0$,$f^{\prime}(x) = \frac{d}{dx} \left( \frac{2x}{4-3x} \right) = \frac{(4-3x)(2) - 2x(-3)}{(4-3x)^2} = \frac{8-6x+6x}{(4-3x)^2} = \frac{8}{(4-3x)^2}$.
Thus,$Lf^{\prime}(0) = \lim_{x \rightarrow 0^-} \frac{8}{(4-3x)^2} = \frac{8}{16} = \frac{1}{2}$.
Since $Lf^{\prime}(0) = Rf^{\prime}(0) = \frac{1}{2}$,the derivative $f^{\prime}(0)$ exists and is equal to $\frac{1}{2}$.

(C) function $f(x)$ is differentiable at $x=a$ if the left-hand derivative $LHD = \lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}$ and right-hand derivative $RHD = \lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$ exist and are equal.
For $f_1(x)=|x|$,at $x=1$,$f_1(x)=x$. Thus $f_1'(1)=1$. It is differentiable.
For $f_2(x)$,at $x=1$,$LHD = \frac{d}{dx}(1+\sin(x-1))|_{x=1} = \cos(0) = 1$. $RHD = \frac{d}{dx}(x)|_{x=1} = 1$. Since $LHD=RHD$,it is differentiable.
For $f_3(x)$,at $x=1$,$LHD = \frac{d}{dx}(x^2+7x-7)|_{x=1} = 2(1)+7 = 9$. $RHD = \frac{d}{dx}(\frac{3x-1}{2})|_{x=1} = \frac{3}{2} = 1.5$. Since $LHD \neq RHD$,$f_3(x)$ is not differentiable at $x=1$.
For $f_4(x)$,at $x=1$,$f_4(x) = |x-1|+|x-2|$. For $x \leq 1$,$f_4(x) = -(x-1)-(x-2) = -2x+3$. $LHD = -2$. For $x > 1$,$f_4(x) = 1+x-x^3$. $RHD = \frac{d}{dx}(1+x-x^3)|_{x=1} = 1-3(1)^2 = -2$. Since $LHD=RHD$,it is differentiable.

(A) To check the differentiability of $f(x)$ at $x = 2$,we first note that for $x$ in a neighborhood of $2$,$\cos(\frac{\pi}{x})$ is positive because $\frac{\pi}{x}$ is near $\frac{\pi}{2}$. Specifically,for $x$ near $2$,$\frac{\pi}{x}$ is in the interval $(0, \pi)$,where $\cos(\frac{\pi}{x})$ is positive.
Thus,for $x$ in a small neighborhood of $2$,$f(x) = x^2 \cos(\frac{\pi}{x})$.
Now,we find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx} [x^2 \cos(\frac{\pi}{x})] = 2x \cos(\frac{\pi}{x}) + x^2 [-\sin(\frac{\pi}{x})] \cdot (-\frac{\pi}{x^2}) = 2x \cos(\frac{\pi}{x}) + \pi \sin(\frac{\pi}{x})$.
Evaluating at $x = 2$:
$f'(2) = 2(2) \cos(\frac{\pi}{2}) + \pi \sin(\frac{\pi}{2}) = 4(0) + \pi(1) = \pi$.
Since the derivative exists at $x = 2$,the function is differentiable at $x = 2$.

(A) The absolute value function can be rewritten as a piecewise function:
$f(x)= \begin{cases} 1-2x, & \text{if } x \leq \frac{1}{2} \\ 2x-1, & \text{if } x > \frac{1}{2} \end{cases}$
Check for continuity at $x=\frac{1}{2}$:
Left-hand limit: $\lim_{x \rightarrow \frac{1}{2}^-} f(x) = \lim_{x \rightarrow \frac{1}{2}} (1-2x) = 1-2(\frac{1}{2}) = 0$.
Right-hand limit: $\lim_{x \rightarrow \frac{1}{2}^+} f(x) = \lim_{x \rightarrow \frac{1}{2}} (2x-1) = 2(\frac{1}{2})-1 = 0$.
Function value: $f(\frac{1}{2}) = |1-2(\frac{1}{2})| = 0$.
Since the left-hand limit,right-hand limit,and function value are all equal,the function is continuous at $x=\frac{1}{2}$.
Check for differentiability at $x=\frac{1}{2}$:
Left-hand derivative: $f'(x) = \frac{d}{dx}(1-2x) = -2$ for $x < \frac{1}{2}$.
Right-hand derivative: $f'(x) = \frac{d}{dx}(2x-1) = 2$ for $x > \frac{1}{2}$.
Since the left-hand derivative $(-2)$ and right-hand derivative $(2)$ are not equal,the function is not differentiable at $x=\frac{1}{2}$.

(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$.
Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$,we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a_3|x|x^2$.
For $f(x)$ to be differentiable at $x=0$,the left-hand derivative and right-hand derivative must be equal.
Right-hand derivative at $x=0$: $f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{a_0 + a_1h + a_2h^2 + a_3h^3 - a_0}{h} = \lim_{h \to 0^+} (a_1 + a_2h + a_3h^2) = a_1$.
Left-hand derivative at $x=0$: $f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{a_0 + a_1(-h) + a_2h^2 + a_3(-h^3) - a_0}{h} = \lim_{h \to 0^-} (-a_1 + a_2h - a_3h^2) = -a_1$.
For differentiability,$f'(0^+) = f'(0^-) \implies a_1 = -a_1 \implies 2a_1 = 0 \implies a_1 = 0$.
Thus,$f(x)$ is differentiable at $x=0$ only if $a_1=0$.

(A) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.
$\text{LHL} = \lim_{x \to 1^-} (x^2 + 3x + a) = 1 + 3 + a = 4 + a$.
$\text{RHL} = \lim_{x \to 1^+} (bx + 2) = b + 2$.
Since $\text{LHL} = \text{RHL}$,we have $4 + a = b + 2$,which implies $b - a = 2$ (Equation $1$).
For differentiability,$\text{LHD} = \text{RHD}$ at $x = 1$.
$\text{LHD} = \frac{d}{dx}(x^2 + 3x + a) = 2x + 3$. At $x = 1$,$\text{LHD} = 2(1) + 3 = 5$.
$\text{RHD} = \frac{d}{dx}(bx + 2) = b$.
Thus,$b = 5$.
Substituting $b = 5$ into Equation $1$: $5 - a = 2 \Rightarrow a = 3$.

(B) The given equation is $x^2 - 3x + 2 = 0$.
Factoring the quadratic equation,we get $(x - 1)(x - 2) = 0$.
Thus,the roots are $\alpha = 1$ and $\beta = 2$.
So,$f(x) = |x - 1| + |x - 2|$.
For $x \in [1, 2]$,we have $x - 1 \ge 0$ and $x - 2 \le 0$.
Therefore,$f(x) = (x - 1) - (x - 2) = x - 1 - x + 2 = 1$.
Since $f(x) = 1$ is a constant function on the interval $[1, 2]$,it is differentiable at every point in the interval $[1, 2]$.
Thus,the number of points in $[1, 2]$ at which $f$ is not differentiable is $0$.

(D) To check the differentiability of $f(x)$ at $x=1$,we calculate the derivative using the definition: $f^{\prime}(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$.
Given $f(1) = 1$ and for $h \neq 0$,$f(1+h) = e^{((1+h)^{10}-1)} + h^2 \sin(\frac{1}{h})$.
Substituting these into the limit definition:
$f^{\prime}(1) = \lim_{h \to 0} \frac{e^{((1+h)^{10}-1)} + h^2 \sin(\frac{1}{h}) - 1}{h}$.
Using the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$,where $u = (1+h)^{10}-1 = 1 + 10h + \dots - 1 = 10h + O(h^2)$:
$f^{\prime}(1) = \lim_{h \to 0} \frac{1 + (10h + O(h^2)) + h^2 \sin(\frac{1}{h}) - 1}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{10h + O(h^2) + h^2 \sin(\frac{1}{h})}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} (10 + O(h) + h \sin(\frac{1}{h}))$.
Since $\lim_{h \to 0} h \sin(\frac{1}{h}) = 0$ by the Squeeze Theorem,we get $f^{\prime}(1) = 10 + 0 + 0 = 10$.

(C) For $f(x) = a \sin |x| + b e^{|x|}$ to be differentiable at $x = 0$,the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ must be equal.
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{a \sin |-h| + b e^{|-h|} - (a \sin 0 + b e^0)}{h} = \lim_{h \to 0} \frac{a \sin(-h) + b e^{-h} - b}{h} = \lim_{h \to 0} \frac{-a \sin h + b(e^{-h} - 1)}{h} = -a - b$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{a \sin h + b e^h - b}{h} = \lim_{h \to 0} \frac{a \sin h + b(e^h - 1)}{h} = a + b$.
For differentiability,$LHD = RHD \implies -a - b = a + b \implies 2a + 2b = 0 \implies a + b = 0$.

(C) We are given $g(x) = x f^{\prime}(x)$.
To find $g^{\prime}(0)$,we use the definition of the derivative:
$g^{\prime}(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}$.
Since $g(0) = 0 \cdot f^{\prime}(0) = 0$,we have:
$g^{\prime}(0) = \lim_{h \to 0} \frac{h f^{\prime}(h) - 0}{h} = \lim_{h \to 0} f^{\prime}(h)$.
Since $f^{\prime}(x)$ is given to be continuous at $x=0$,we have $\lim_{h \to 0} f^{\prime}(h) = f^{\prime}(0)$.
Given $f^{\prime}(0) = 1$,it follows that $g^{\prime}(0) = 1$.

(C) We are given the function $f(x) = \max \{a-x, a+x, b\}$.
To find the points of non-differentiability,we analyze the intersections of the functions $y_1 = a-x$,$y_2 = a+x$,and $y_3 = b$.
$1$. Intersection of $y_1$ and $y_3$: $a-x = b \implies x = a-b$.
$2$. Intersection of $y_2$ and $y_3$: $a+x = b \implies x = b-a$.
$3$. Intersection of $y_1$ and $y_2$: $a-x = a+x \implies 2x = 0 \implies x = 0$.
Since $0 < a < b$,we have $a-b < 0$ and $b-a > 0$.
At $x = a-b$,$f(x)$ transitions from $a-x$ to $b$,creating a sharp turn.
At $x = b-a$,$f(x)$ transitions from $b$ to $a+x$,creating a sharp turn.
At $x = 0$,the value of $f(0) = \max \{a, a, b\} = b$ (since $b > a$). The function is $b$ in the interval $[a-b, b-a]$,so it is differentiable at $x=0$.
Thus,there are exactly $2$ points of non-differentiability: $x = a-b$ and $x = b-a$.

(B) For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$ and $f'(1^-) = f'(1^+)$.
First,for continuity at $x = 1$,we have $f(1^-) = f(1^+)$.
$f(1^-) = \lim_{x \to 1^-} (2 \alpha (x^2 - 2) + 2 \beta x) = 2 \alpha (1 - 2) + 2 \beta (1) = -2 \alpha + 2 \beta$.
$f(1^+) = (\alpha + 3)(1) + (\alpha - \beta) = 2 \alpha - \beta + 3$.
Equating them: $-2 \alpha + 2 \beta = 2 \alpha - \beta + 3 \Rightarrow 4 \alpha - 3 \beta = -3$ $\quad (1)$.
Next,for differentiability,$f'(1^-) = f'(1^+)$.
$f'(x) = 4 \alpha x + 2 \beta$ for $x < 1$ and $f'(x) = \alpha + 3$ for $x > 1$.
$f'(1^-) = 4 \alpha + 2 \beta$.
$f'(1^+) = \alpha + 3$.
Equating them: $4 \alpha + 2 \beta = \alpha + 3 \Rightarrow 3 \alpha + 2 \beta = 3$ $\quad (2)$.
Solving equations $(1)$ and $(2)$:
From $(2)$,$2 \beta = 3 - 3 \alpha \Rightarrow \beta = \frac{3 - 3 \alpha}{2}$.
Substitute into $(1)$: $4 \alpha - 3(\frac{3 - 3 \alpha}{2}) = -3 \Rightarrow 8 \alpha - 9 + 9 \alpha = -6 \Rightarrow 17 \alpha = 3 \Rightarrow \alpha = \frac{3}{17}$.
Then $\beta = \frac{3 - 3(3/17)}{2} = \frac{3 - 9/17}{2} = \frac{42/17}{2} = \frac{21}{17}$.
Thus,$34(\alpha + \beta) = 34(\frac{3}{17} + \frac{21}{17}) = 34(\frac{24}{17}) = 2 \times 24 = 48$.

(D) The function is $f(x) = e^{\sin |x|} - |x|$.
First, check the differentiability at $x = 0$. The function $f(x)$ contains the term $|x|$, which is not differentiable at $x = 0$. Therefore, $f(x)$ is not differentiable at $x = 0$. Thus, Statement $I$ is false.
Next, consider Statement $II$ in the interval $(-\pi, -\frac{\pi}{2})$. For $x < 0$, we have $|x| = -x$. Thus, $f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x$.
The derivative is $f'(x) = e^{-\sin x}(-\cos x) + 1 = 1 - e^{-\sin x}\cos x$.
For $x \in (-\pi, -\frac{\pi}{2})$, we have $\sin x \in (-1, 0)$ and $\cos x \in (-1, 0)$.
Since $\sin x$ is negative, $-\sin x$ is positive, so $e^{-\sin x} > e^0 = 1$. Also, $\cos x$ is negative.
Thus, $-e^{-\sin x}\cos x$ is positive. Therefore, $f'(x) = 1 + (\text{positive value}) > 0$.
Since $f'(x) > 0$, the function $f(x)$ is strictly increasing in the interval $(-\pi, -\frac{\pi}{2})$. Thus, Statement $II$ is true.